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Math Help - Physics problem

  1. #1
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    Talking Physics problem

    A cross country skier skis is 2.80 km in the direction 45.0˚ west of south, then 7.40 km in the direction 30.0˚ north of east, and finally 3.30 km in the direction 22.0˚ south of west. A) Show these displacements in a diagram. B) How far is the skier from the starting point?
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  2. #2
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    Quote Originally Posted by jasonlewiz View Post
    A cross country skier skis is 2.80 km in the direction 45.0˚ west of south, then 7.40 km in the direction 30.0˚ north of east, and finally 3.30 km in the direction 22.0˚ south of west. A) Show these displacements in a diagram. B) How far is the skier from the starting point?
    let the skier start at the origin. use components to find the displacement in the x and y directions ...

    r_x = -2.8\cos(45) + 7.4\cos(30) - 3.3\cos(22)

    r_y = -2.8\sin(45) + 7.4\sin(30) - 3.3\sin(22)

    |r| = \sqrt{(r_x)^2 + (r_y)^2}
    Last edited by skeeter; November 21st 2009 at 07:08 AM. Reason: fixed
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  3. #3
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    Quote Originally Posted by skeeter View Post
    let the skier start at the origin. use components to find the displacement in the x and y directions ...

    r_x = -2.8\cos(45) + 7.4\cos(30) - 3.3\cos(22)

    r_y = -2.8\sin(45) + 7.4\cos(30) - 3.3\cos(22)

    |r| = \sqrt{(r_x)^2 + (r_y)^2}
    7.4\cos(30) for both x&y displacements & -3.3\cos(22) for both x&y displacements

    I'll bet you did that just to see if we were awake?

    .
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  4. #4
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    no ... I'm not awake yet

    r_x = -2.8\cos(45) + 7.4\cos(30) - 3.3\cos(22)

    r_y = -2.8\sin(45) + 7.4\sin(30) - 3.3\sin(22)

    |r| = \sqrt{(r_x)^2 + (r_y)^2}

    copied and only changed one , sorry.
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